Gas Volume Calculations

Gas volume calculations use the molar volume to convert between moles of a gas and its volume. At room temperature and pressure (RTP), one mole of any gas occupies 24 dm3. This single fact unlocks a wide range of stoichiometry questions involving gases.

The method

1. Write a balanced equation if the question involves a reaction.
2. Find the moles of gas involved (either given directly or calculated from mass, concentration, or the equation).
3. Use the molar volume relationship: volume = moles x 24 dm3 (at RTP).
4. Convert units if necessary (1 dm3 = 1000 cm3).

The formula

volume (dm3) = moles x 24

moles = volume (dm3) / 24

Where:

• Volume is in dm3 at RTP
• 24 dm3/mol is the molar volume at RTP

If volume is in cm3: volume (cm3) = moles x 24000 and moles = volume (cm3) / 24000

Worked examples

Example 1: Volume from moles

Question: Calculate the volume of 0.5 mol of carbon dioxide at RTP.

Working:

• Volume = moles x 24 = 0.5 x 24 = 12 dm3

Answer: 12 dm3

Examiner note: Direct substitution. State the volume in dm3 unless the question asks for cm3.

Example 2: Moles from volume

Question: How many moles of oxygen are present in 4800 cm3 of gas at RTP?

Working:

• Convert: 4800 cm3 = 4.8 dm3
• Moles = 4.8 / 24 = 0.2 mol

Answer: 0.2 mol

Examiner note: Convert cm3 to dm3 first, or divide by 24000 directly.

Example 3: Volume of gas from a reaction

Question: What volume of hydrogen gas (at RTP) is produced when 2.4 g of magnesium reacts with excess hydrochloric acid? (Ar: Mg = 24)

Working:

• Balanced equation: Mg + 2HCl -> MgCl2 + H2
• Moles of Mg = 2.4 / 24 = 0.1 mol
• Mole ratio: 1 Mg : 1 H2
• Moles of H2 = 0.1 mol
• Volume of H2 = 0.1 x 24 = 2.4 dm3 (or 2400 cm3)

Answer: 2.4 dm3

Examiner note: This combines reacting masses with gas volume. The sequence is mass -> moles -> mole ratio -> moles of gas -> volume.

Example 4: Volume of gas from thermal decomposition

Question: What volume of carbon dioxide (at RTP) is produced when 5 g of calcium carbonate is heated? (Ar: Ca = 40, C = 12, O = 16)

Working:

• Balanced equation: CaCO3 -> CaO + CO2
• Mr of CaCO3 = 100
• Moles of CaCO3 = 5 / 100 = 0.05 mol
• Mole ratio: 1:1
• Moles of CO2 = 0.05 mol
• Volume of CO2 = 0.05 x 24 = 1.2 dm3 (or 1200 cm3)

Answer: 1.2 dm3

Examiner note: The 1:1 ratio makes this straightforward. Always show the mole ratio step.

Example 5: Mass from gas volume

Question: 480 cm3 of chlorine gas at RTP reacts with sodium. What mass of sodium chloride is produced? (Ar: Na = 23, Cl = 35.5)

Working:

• Balanced equation: 2Na + Cl2 -> 2NaCl
• Moles of Cl2 = 480 / 24000 = 0.02 mol
• Mole ratio: 1 Cl2 : 2 NaCl
• Moles of NaCl = 0.02 x 2 = 0.04 mol
• Mr of NaCl = 23 + 35.5 = 58.5
• Mass of NaCl = 0.04 x 58.5 = 2.34 g

Answer: 2.34 g

Examiner note: This goes in the reverse direction: volume -> moles -> mole ratio -> moles -> mass. The mole ratio is 1:2, so double the moles.

Common mistakes

1. Using the wrong molar volume. At RTP, use 24 dm3/mol. At STP, use 22.4 dm3/mol. Check what the question states.
2. Mixing up dm3 and cm3. 1 dm3 = 1000 cm3. Use 24 for dm3 and 24000 for cm3.
3. Applying molar volume to solids or liquids. The 24 dm3/mol value applies only to gases. Do not use it for solids or liquids.
4. Forgetting the mole ratio. When the question involves a reaction, you must use the balanced equation to convert between moles of different substances.

When this appears

Gas volume calculations are Extended tier content and appear on Paper 4. They build on moles from mass and reacting masses. See the stoichiometry topic for context.